Sure, let's solve these two equations step-by-step.
### First Equation: [tex]\((x + 3)^2 = 6x + 18\)[/tex]
1. Expand the left side of the equation:
[tex]\[
(x + 3)^2 = (x + 3)(x + 3) = x^2 + 6x + 9
\][/tex]
Now the equation becomes:
[tex]\[
x^2 + 6x + 9 = 6x + 18
\][/tex]
2. Simplify the equation:
Subtract [tex]\(6x + 18\)[/tex] from both sides:
[tex]\[
x^2 + 6x + 9 - 6x - 18 = 0
\][/tex]
Simplify it:
[tex]\[
x^2 - 9 = 0
\][/tex]
3. Solve the quadratic equation:
Add 9 to both sides:
[tex]\[
x^2 = 9
\][/tex]
Take the square root of both sides:
[tex]\[
x = \pm 3
\][/tex]
Thus, the solutions are:
[tex]\[
x = 3 \quad \text{or} \quad x = -3
\][/tex]
### Second Equation: [tex]\(16x^2 - 1 = 0\)[/tex]
1. Isolate [tex]\(x^2\)[/tex]:
Add 1 to both sides:
[tex]\[
16x^2 = 1
\][/tex]
2. Solve for [tex]\(x^2\)[/tex]:
Divide both sides by 16:
[tex]\[
x^2 = \frac{1}{16}
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
Take the square root of both sides:
[tex]\[
x = \pm \frac{1}{4}
\][/tex]
Thus, the solutions are:
[tex]\[
x = \frac{1}{4} \quad \text{or} \quad x = -\frac{1}{4}
\][/tex]
### Summary of Solutions:
- For the equation [tex]\((x + 3)^2 = 6x + 18\)[/tex], the solutions are [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex].
- For the equation [tex]\(16x^2 - 1 = 0\)[/tex], the solutions are [tex]\(x = \frac{1}{4}\)[/tex] and [tex]\(x = -\frac{1}{4}\)[/tex].
Putting it all together, the complete set of solutions is:
[tex]\[
x = -3, \quad x = 3, \quad x = -\frac{1}{4}, \quad x = \frac{1}{4}
\][/tex]
